Substituindo na série e escrevendo os primeiros termos, temos:
sin x = 12 + raiz32,(x-tfracpi6) - 12,(x-tfracpi6)^22! - raiz32,(x-tfracpi6)^33! + 12,(x-tfracpi6)^44! + cdots
Passo 5
Em forma de soma agrupando o período 4, a série fica
sin x = sumk=0^infty gl[ 12,(x-tfracpi6)^4k(4k)! + raiz32,(x-tfracpi6)^4k+1(4k+1)! - 12,(x-tfracpi6)^4k+2(4k+2)! - raiz32,(x-tfracpi6)^4k+3(4k+3)! gr]
Resposta
sin x = 12 + raiz32(x-tfracpi6) - 12,(x-tfracpi6)^22! - raiz32,(x-tfracpi6)^33! + 12,(x-tfracpi6)^44! + cdots
ou sin x = sumk=0^infty l[ tfrac12,tfrac(x-tfracpi6)^4k(4k)! + tfracraiz32,tfrac(x-tfracpi6)^4k+1(4k+1)! - tfrac12,tfrac(x-tfracpi6)^4k+2(4k+2)! - tfracraiz32,tfrac(x-tfracpi6)^4k+3(4k+3)! r]